Let's break down the problem into two parts:
### Part A:
To find the probability that a customer who bought an item at the regular price also purchased the extended warranty, we will employ conditional probability. The conditional probability [tex]\(P(A|B)\)[/tex] is given by the formula:
[tex]\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \][/tex]
In this context:
- [tex]\(A\)[/tex] is the event that a customer purchased an extended warranty.
- [tex]\(B\)[/tex] is the event that a customer bought an item at the regular price.
- [tex]\(P(A \cap B)\)[/tex] is the joint probability that a customer bought an item at the regular price and purchased an extended warranty (0.21).
- [tex]\(P(B)\)[/tex] is the total probability that a customer bought an item at the regular price, which is the sum of the probabilities of buying at the regular price, with and without the warranty.
First, we compute [tex]\(P(B)\)[/tex]:
[tex]\[ P(B) = P(\text{Regular Price and Warranty}) + P(\text{Regular Price and No Warranty}) \][/tex]
[tex]\[ P(B) = 0.21 + 0.56 \][/tex]
[tex]\[ P(B) = 0.77 \][/tex]
Now, we calculate the conditional probability [tex]\(P(A|B)\)[/tex]:
[tex]\[ P(\text{Warranty | Regular Price}) = \frac{P(\text{Regular Price and Warranty})}{P(\text{Regular Price})} \][/tex]
[tex]\[ P(\text{Warranty | Regular Price}) = \frac{0.21}{0.77} \][/tex]
[tex]\[ P(\text{Warranty | Regular Price}) \approx 0.2727 \][/tex]
So, the probability [tex]\(= \boxed{0.2727}\)[/tex].
### Part B:
To find the probability that a customer buys an extended warranty, we need to sum the probabilities of all cases where the customer buys the warranty.
Here, we are looking for the total probability of purchasing an extended warranty, regardless of the price at which the item was bought.
[tex]\[ P(\text{Warranty}) = P(\text{Regular Price and Warranty}) + P(\text{Sale Price and Warranty}) \][/tex]
[tex]\[ P(\text{Warranty}) = 0.21 + 0.11 \][/tex]
[tex]\[ P(\text{Warranty}) = 0.32 \][/tex]
So, the probability [tex]\(= \boxed{0.32}\)[/tex].
